Can't answer your questions, but I can say that the inverse square law can't be expected to be valid between two points at ground level. The existence of the ground, terrain, trees, structures, etc. will screw it up.
–
garypJul 25 '14 at 13:46

1

Also, the temperature gradient in the vertical direction is known to affect the propagation of sound waves significantly.
–
auxsvrJul 25 '14 at 13:59

1

I thought it was half the intensity, minus three decibels.
–
BMSJul 25 '14 at 15:22

@BMS: Yes, it looks like you are correct. Thank you.
–
dotancohenJul 25 '14 at 15:27

1

@ChrisWhite: the first reference gives the intensity of a 1000 lb bomb at 250 feet which is around 75 m.
–
EdwardJul 25 '14 at 15:40

3 Answers
3

The problem is with your first calculation and also with the somewhat misleading equation that you've found. It's true that $$\frac{I_2}{I_1}=\left(\frac{d_1}{d_2}\right)^2$$ but units are important here. In that formula, $I_1$ and $I_2$ would properly be expressed as power values. To compute with decibels, which are logarithmic quantities, one would instead use $$I_1 + 10\log\left(\frac{d_1}{d_2}\right)^2 = I_2$$ or equivalently, $$I_1 + 20\log\left(\frac{d_1}{d_2}\right) = I_2$$, where $I_1$ and $I_2$ are decibels and $d_1$ and $d_2$ are in identical linear units (feet or meters, for example).

Thank you Edward. Considering then the facts as you've stated them, a 1 ton bomb dropped 45 KM away is as loud as an NHRA Dragster right next to someone? That just does not seem realistic!
–
dotancohenJul 25 '14 at 15:22

2

@dotancohen: It isn't realistic. On the surface of the earth, propagation of sound is affected by many other factors, including the frequencies of the sound, attenuation due to objects between you and the sound, multipath effects from echos and many others. You may find this wikibook of interest.
–
EdwardJul 25 '14 at 15:34

I see now where my confusion was. Thank you for showing the maths explained, and especially for the comment which explains why the perceived value to my ears is so much lower that what the maths make us expect.
–
dotancohenJul 25 '14 at 20:05

I'll piece together some of what's been said in answers and comments in a different light.

With acoustics, it pays to be very careful with units. A sound wave has a pressure $p$, and this corresponds to what I'll call it's intensity $I$. Intensity goes as the square of pressure:
$$ \frac{I}{I_0} = \left(\frac{p}{p_0}\right)^2. $$
Here $I_0$ is the threshold of human hearing, corresponding to pressure $p_0 = 20\ \mathrm{μPa}$. Intensity also falls off as the square of distance:
$$ \frac{I_1}{I_2} = \left(\frac{d_2}{d_1}\right)^2. $$

One catch is that sound has problems traveling long distances near the ground. Things get in the way, and its propagation is complicated. I won't attempt to model that here, but it probably means we're overestimating the sound far from the source.

Another point important for this particular problem is that different measurements will fold in different types of energy. In particular, this chart says a $1$-ton bomb at $250$ feet comes in at $213\ \mathrm{dB}$, including all power such as wind. The sound (pressure wave) is only $176\ \mathrm{dB}$ at the same distance.

Another catch is that human perception doesn't quite follow the loudness scale. Really one should correct for our ears being more attuned to certain frequencies than others. One way to do this is ask, "How loud will a pure $1\ \mathrm{kHz}$ sound need to be to sound as loud to our ears?" The result is the value $L_N$ in phons. Since we don't have any frequency data, I'll just skip this step, but in reality low rumblings don't sound as loud as their raw loudness $L$ values imply. For very low frequencies the difference can be dramatic.

Even once different frequencies are accounted for, our hearing still doesn't quite jibe with decibels. A better measurement is done in sones, leading to a perceived loudness $N$. The formula is
$$ N = \left(10^{L_N/(10\ \mathrm{dB})-4}\right)^{0.30103}. $$

At $75\ \mathrm{m}$ we have $L_1 = 176\ \mathrm{dB}$, so we assume $L_{N,1} = 176\ \mathrm{dB}$, meaning $N_1 = 12{,}400\ \mathrm{sone}$. At $45\ \mathrm{km}$ we have $L_2 = L_{N,2} = 120\ \mathrm{dB}$, meaning $N_2 = 256\ \mathrm{sone}$. That puts it at the threshold of hearing damage even for short exposures, but remember the actual value is probably much lower due to all the sound being in low frequencies.

Thank you. You certain understand how to phrase the ideas into terms that are easier to digest, and to address the points of confusion.
–
dotancohenJul 25 '14 at 20:03

@ChrisWhite: you've mentioned phons and sones but have neglected scones which is defined as "the maximum sound level that a typical adult human can be subjected to during the preparation of breakfast tea and still plausibly pretend to be asleep." :)
–
EdwardJul 25 '14 at 20:13

Thank you. The online information that I've been able to locate has been loose in their interpretations and strict in their expectations of the level of knowledge of the reader. Thank you for stating it so clearly!
–
dotancohenJul 25 '14 at 15:20